The Advatage Testig Foudatio 202 Problem I the morig, Esther biked from home to school at a average speed of x miles per hour. I the afteroo, havig let her bike to a fried, Esther walked back home alog the same route at a average speed of 3 miles per hour. Her average speed for the roud trip was 5 miles per hour. What is the value of x? Aswer: 5. Solutio: Let d be the umber of miles betwee home ad school. Goig from home to school takes d hours. Goig from school to home takes d hours. x 3 The etire roud trip takes 2d hours. So we get the equatio 5 Dividig by d gives Solvig for x gives x = 5. d x + d 3 = 2d 5. x + 3 = 2 5. Problem 2 I the figure below, the ceters of the six cogruet circles form a regular hexago with side legth 2.
Adjacet circles are taget to each other. What is the area of the shaded regio? Express your aswer i the form a bπ, where a ad b are positive itegers ad is a square-free positive iteger. Aswer: 6 3 2π. Solutio: We ca split the hexago ito 6 equilateral triagles with side legth 2. Sice the area of a equilateral triagle with side legth x is x2 3, 4 the area of each of the 6 equilateral triagles is 3. So the area of the hexago is 6 3. The 6 circle sectors altogether form 2 full circles of radius. So the combied area of the sectors is 2π. Thus the area of the shaded regio is 6 3 2π. Problem 3 What is the least positive iteger such that! is a multiple of 202 202? Aswer:,00,527. Solutio: The prime factorizatio of 202 is 2 2 503. So the prime factorizatio of 202 202 is (2 2 503 202 = 2 4024 503 202. Thus! is a multiple of 202 202 if ad oly if it has at least 202 factors of 503 ad at least 4024 factors of 2. If < 2009 503, the the umber of factors of 503 i! is + 2008 + 3 = 20 < 202. 503 503 2 So we eed 2009 503. I that case, the umber of factors of 503 i! is at least + 2009 + 3 = 202. 503 503 2 Also, i that case, the umber of factors of 2 i! is at least 2009 503 500,000 > 4024. 2 2 So the least value of that works is 2009 503 = (2000+9(500+3 =,000,000+4500+6000+27 =,00,527. 2
Problem 4 Evaluate the expressio 2 ( ( 3 7 + 69 Aswer: 4. ( 3 7 + 3 ( 7 7 + 289 ( + 7 ( 3 3. Solutio: Let a =, let b = 3, ad let c = 7. Usig these variables, the expressio becomes a ( 2 ( b c + b 2 ( c a + c 2 a b a ( ( b c + b ( c a + c. a b By groupig the terms i the umerator ad deomiator with the same reciprocal, we ca rewrite this expressio as a (c2 b 2 + b (a2 c 2 + c (b2 a 2 (c b + (a c + (b a. a b c Usig differece of squares, ad lettig S = a + b + c, the umerator of this fractio ca be rewritte as follows: a (c b(c + b + b (a c(a + c + (b a(b + a c = a (c b(s a + b (a c(s b + (b a(s c c = a (c bs (c b + b (a cs (a c + (b as (b a [ c = a (c b + b (a c + ] (b a S. c But this fial expressio is the deomiator of our fractio times S. fractio simplifies to just S. The value of the fractio is So S = a + b + c = + 3 + 7 = 4. Problem 5 The figure below shows a semicircle iscribed i a right triagle. 3
The triagle has legs of legth 8 ad 5. The semicircle is taget to the two legs, ad its diameter is o the hypoteuse. What is the radius of the semicircle? Express your aswer as a fractio i simplest form. Aswer: 20 23. Solutio: (0, 8 r (r, r r (0, 0 (5, 0 I the figure above, we have draw radii to the two poits where the semicircle is taget to the legs. Let r be the radius of the semicircle. We have imposed a coordiate system o the figure. The hypoteuse has edpoits (5, 0 ad (0, 8. Usig the itercept-itercept form, we see that the hypoteuse lies o the lie x + y =. 5 8 The ceter of the semicircle is (r, r. Because the ceter is o the hypoteuse, we have r 5 + r 8 So r = 20 23. =. We ca add the two fractios to get 23r 20 =. Problem 6 For how may ordered pairs of positive itegers (x, y is the least commo multiple of x ad y equal to,003,003,00? Aswer: 343. Solutio: Let s first factorize,003,003,00 ito primes:,003,003,00 = 00 3 = (7 3 3 = 7 3 3 3 3. 4
For the least commo multiple (lcm of x ad y to have this factorizatio, x must have the form 7 a a 2 3 a 3 ad y must have the form 7 b b 2 3 b 3. We the have 7 3 3 3 3 = lcm(x, y = 7 max(a,b max(a 2,b 2 3 max(a 3,b 3. So max(a, b = max(a 2, b 2 = max(a 3, b 3 = 3. To make max(a, b = 3, there are 7 choices for (a, b : (0, 3, (, 3, (2, 3, (3, 3, (3, 2, (3,, (3, 0. Similarly, there are 7 choices for (a 2, b 2 ad 7 choices for (a 3, b 3. Altogether, the umber of choices is 7 7 7 = 343. Problem 7 Let f, f 2, f 3,..., be a sequece of umbers such that f = f + f 2 for every iteger 3. If f 7 = 83, what is the sum of the first 0 terms of the sequece? Aswer: 93. Solutio: Let s express each of the first 0 terms usig oly f ad f 2 : f = f f 2 = f 2 f 3 = f 2 + f f 4 = 2f 2 + f f 5 = 3f 2 + 2f f 6 = 5f 2 + 3f f 7 = 8f 2 + 5f f 8 = 3f 2 + 8f f 9 = 2f 2 + 3f f 0 = 34f 2 + 2f. (Did you otice that the coefficiets are Fiboacci umbers? Addig both sides, we see that the sum of the first 0 terms is equal to 88f 2 + 55f. Therefore, the sum of the first 0 terms is 88f 2 + 55f = (8f 2 + 5f = f 7 = 83 = 93. 5
Problem 8 Suppose that x, y, ad z are real umbers such that x+y+z = 3 ad x 2 + y 2 + z 2 = 6. What is the largest possible value of z? Express your aswer i the form a + b, where a ad b are positive itegers. Aswer: + 2. Solutio: The two equatios ca be rewritte as x+y = 3 z ad x 2 +y 2 = 6 z 2. We ca get a iequality for z as follows: 2(6 z 2 = 2(x 2 + y 2 = (x + y 2 + (x y 2 (x + y 2 = (3 z 2. Expadig, we get Groupig like terms gives ad so 2 2z 2 = 2(6 z 2 (3 z 2 = z 2 6z + 9. By completig the square, we get 3z 2 6z 3, z 2 2z. (z 2 = z 2 2z + + = 2. Takig the square root of both sides gives z 2. So z + 2. To see that z = + 2 is possible, we ca choose x = y = 3 z. So the 2 largest possible value of z is + 2. Problem 9 Biaca has a rectagle whose legth ad width are distict primes less tha 00. Let P be the perimeter of her rectagle, ad let A be the area of her rectagle. What is the least possible value of P 2? Express A your aswer as a fractio i simplest form. Aswer: 82944 583. 6
Solutio: Let l be the legth of the rectagle ad w be the width. Without loss of geerality, assume that l > w. The perimeter is P = 2l + 2w. The area is A = lw. So we have P 2 A = (2l + 2w2 lw = 4l2 + 8lw + 4w 2 lw ( l = 8 + 4 w + w. l To miimize this expressio, we wat l to be as small as possible. w If l ad w are apart, the l = 3 ad w = 2. I that case, the ratio l w is 3 2 İf l ad w are 2 apart, the they are twi primes. The largest pair of twi primes less tha 00 is 73 ad 7. So the ratio l 73 is at least. w 7 If l ad w are at least 3 apart, the l w w + 3 w = + 3 w > + 3 00 > + 2 7 = 73 7. Altogether, we have show that l 73, with equality if l = 73 ad w 7 w = 7. So we have ( P 2 l A = 8 + 4 w + w ( 73 8 + 4 l 7 + 7 [2(73 + 7]2 =. 73 73 7 The umerator is [2(73 + 7] 2 = (2 44 2 = (2 5 3 2 2 = 2 0 3 4 = 024 8 = 82,944. The deomiator is 73 7 = 583. So we have P 2 A 82,944 583, with equality if l = 73 ad w = 7. Hece the least possible value of P 2 is A 82,944 583. Note: This problem was proposed by Oleg Kryzhaovsky. Problem 0 Let ABC be a triagle with a right agle ABC. Let D be the midpoit of BC, let E be the midpoit of AC, ad let F be the midpoit of AB. Let G be the midpoit of EC. Oe of the agles of DF G is a right agle. What is the least possible value of BC? Express your aswer as a AG fractio i simplest form. Aswer: 2 3. 7
Solutio: A(0, 4q F (0, 2q E(2p, 2q G(3p, q B(0, 0 D(2p, 0 C(4p, 0 I the figure above, we have imposed a coordiate system. We have placed poit B at (0, 0, poit C at (4p, 0, ad poit A at (0, 4q, with p ad q positive. (We iserted the factors of 4 to avoid fractios i the coordiates of the other poits. Because D is the midpoit of BC, it is at (2p, 0. Because E is the midpoit of AC, it is at (2p, 2q. Because F is the midpoit of AB, it is at (0, 2q. Because G is the midpoit of EC, it is at (3p, q. The legth of BC is 4p. The legth of AG is 3 p 2 + q 2. The desired ratio is BC AG = 4p 3 p 2 + q 2 = 4 3 + (q/p 2. This ratio is a decreasig fuctio of q. p The slope of DF is q q, the slope of F G is, ad the slope of DG is q. p 3p p Hece the slope of a lie perpedicular to DG is p, the egative reciprocal q of q p Ẇe are give that oe of the agles of DF G is a right agle. Because both DF ad F G have egative slopes, they ca t be perpedicular to each other. Hece either DF or F G is perpedicular to DG. By our previous paragraph o slopes, we have either q = p or q = p. So q is either p q 3p q p or 3. (All the steps above are reversible, so both choices are geometrically realizable. To make the desired ratio BC as small as possible, we should select the AG 8
bigger choice, 3, for q. I that case, p BC AG = 4 3 + (q/p = 4 2 3 + 3 = 2 3. Note: This problem was proposed by Oleg Kryzhaovsky. Problem Aliso has a aalog clock whose hads have the followig legths: a iches (the hour had, b iches (the miute had, ad c iches (the secod had, with a < b < c. The umbers a, b, ad c are cosecutive terms of a arithmetic sequece. The tips of the hads travel the followig distaces durig a day: A iches (the hour had, B iches (the miute had, ad C iches (the secod had. The umbers A, B, ad C (i this order are cosecutive terms of a geometric sequece. What is the value of B A? Express your aswer i the form x + y, where x ad y are positive itegers ad is a square-free positive iteger. Aswer: 60 + 24 5. Solutio: I oe day, the hour had makes 2 revolutios, so A = 2(2πa = 4πa. The miute had makes 24 revolutios, so B = 24(2πb = 48πb. The secod had makes 24 60 = 440 revolutios, so C = 440(2πc = 2880πc. Let r be the commo ratio of the geometric sequece A, B, ad C. Because B = Ar, we have 48πb = 4πar, which simplifies to 2b = ar. Because a < b, we deduce that r > 2. Because C = Ar 2, we have 2880πc = (4πar 2, which simplifies to 720c = ar 2. Because a, b, ad c form a arithmetic sequece, we have a + c = 2b. By the previous paragraph, we have a + 720 ar2 = 6 ar, which simplifies to r 2 20r + 720 = 0. By the quadratic formula, we have r = 20 ± 20 2 4(720 2 = 60 ± 60 2 720 = 60 ± 24 5. Because r > 2, we have r = 60 + 24 5. Note: This problem was proposed by Oleg Kryzhaovsky. 9
Problem 2 What is the sum of all positive iteger values of that satisfy the equatio ( π ( 2π ( 4π ( 8π ( 6π cos cos cos cos cos = 32? Aswer: 47. Solutio: This problem ivolves the cosie of agles i a doublig sequece. That suggests usig a double-agle idetity such as si 2θ = 2 si θ cos θ. To use that idetity, let s multiply both sides of the give equatio by si π : ( π ( π ( 2π ( 4π ( 8π ( 6π si cos cos cos cos cos = ( π 32 si. We ca collapse the left-had side by usig the double-agle idetity may times: ( π ( π ( 2π ( 4π ( 8π ( 6π si cos cos cos cos cos = ( 2π ( 2π ( 4π ( 8π ( 6π 2 si cos cos cos cos = ( 4π ( 4π ( 8π ( 6π 4 si cos cos cos = ( 8π ( 8π ( 6π 8 si cos cos = ( 6π ( 6π 6 si cos = ( 32π 32 si. So our equatio becomes ( 32π 32 si = ( π 32 si. We ca remove the factors of o both sides: 32 ( 32π ( π si = si. 0
Oe way to proceed is to use the sum-to-product idetity ( α β ( α + β si α si β = 2 si cos. 2 2 Applyig this idetity with α = 32π ad β = π, we get the equatio ( 3π ( 33π 2 si cos = 0. 2 2 So either si 3π 33π or cos is 0. 2 2 If si 3π 3π = 0, the is a multiple of π. So 3 is a multiple of 2. But 2 2 3 is odd, so this case has o solutios. If cos 33π 33π = 0, the is π plus a multiple of π. So 33 is a odd umber 2 2 2 times. That meas is either, 3,, or 33. Puttig both cases together, we see that is either, 3,, or 33. All the steps above are reversible, except for multiplyig both sides of the origial equatio by si π. If >, the this factor is positive, but if =, the this factor is zero. Ideed, whe we plug = ito the origial equatio, we see that it is ot a solutio. Hece there are exactly three solutios to the origial equatio: 3,, ad 33. Their sum is 3 + + 33 = 47. Problem 3 For how may itegers with 202 is the product ( ( + e 2πik/ + equal to zero? Aswer: 335. k=0 Solutio: If the product is zero, the oe of the factors ( + e 2πik/ + is zero. I that case, ( + e 2πik/ is. That implies + e 2πik/ has absolute value. So the three complex umbers, 0, ad e 2πik/ are the vertices of a equilateral triagle with side legth. Thus e 2πik/ is either e 2πi/3 or its cojugate. I that case, + e 2πik/ is either e πi/3 or its cojugate. So ( + e 2πik/ is either e πi/3 or its cojugate. The oly way that ca be is if is a odd multiple of 3. (Ad if is a odd multiple of 3, the the factor correspodig to k = /3 will be zero. So our problem boils dow to coutig the odd multiples of 3 betwee ad 202. There are 670 multiples of 3 i this iterval. Thus the umber of odd multiples of 3 i this iterval is half that umber, or 335.
Problem 4 Let k be the smallest positive iteger such that the biomial coefficiet ( ( 0 9 k is less tha the biomial coefficiet 0 9 + k. Let a be the first (from the left digit of k ad let b be the secod (from the left digit of k. What is the value of 0a + b? Aswer: 38. Solutio: Let = 0 9 +. Our iequality is ( ( <. k k We will use the followig two idetities o biomial coefficiets: ( ( = ( k k k ( ( ( k + = k. k k Combiig these two idetities gives a third idetity ( ( k = ( k + ( k. k k So our desired iequality is equivalet to ( k + ( k < k. To get a feelig for this iequality, let s temporarily igore the + ad chage the iequality to a equatio: ( k 2 = k. This equatio is quadratic i k, ad so ca be solved by the quadratic formula. The smaller root is k = α, where α = 3 5 is the smaller root of 2 ( α 2 = α. Because 5 is betwee 2.23 ad 2.24, the value of α is betwee 0.38 ad 0.385. So we expect the smallest iteger k that satisfies the iequality starts with 38. We will verify this claim rigorously below. First, we will show that if k α, the the iequality fails: ( k + ( k ( k 2 ( α 2 2 = α 2 k. 2
Secod, we will show that if k α( +, the the iequality holds: ( k + ( k < ( α( + ( α = α( + k. So the smallest positive iteger k that satisfies the iequality is betwee α ad α( + +. Hece a (the first digit of k is ideed 3, ad b (the secod digit of k is 8. Therefore 0a + b is 38. Problem 5 Kate has two bags X ad Y. Bag X cotais 5 red marbles (ad othig else. Bag Y cotais 4 red marbles ad blue marble (ad othig else. Kate chooses oe of her bags at radom (each with probability ad removes a radom marble from that bag (each marble i that 2 bag beig equally likely. She repeats the previous step util oe of the bags becomes empty. At that poit, what is the probability that the blue marble is still i bag Y? Express your aswer as a fractio i simplest form. 63 Aswer: 256. Solutio: Let k be a iteger betwee ad 5 (iclusive. We will first compute the probability that at the ed there are exactly k marbles i bag Y. The fial marble removed must have bee from bag X. Before that, 4 marbles were removed from bag X ad 5 k marbles were removed from bag Y. There are ( 9 k 4 ways of choosig the bag sequece. The probability of makig a particular choice is (/2 0 k. So the probability that at the ed there are exactly k marbles i bag Y is ( 9 k 4 2 k. 2 0 We ca use this result to fid the expected value of the umber of marbles remaiig i bag Y. The expected value is 2 0 5 ( 9 k k= 4 2 k k. Let s compute the sum: 5 ( ( ( ( ( ( 9 k 8 7 6 5 4 2 k k = 2 + 8 + 24 + 64 + 60 4 4 4 4 4 4 k= = 2 70 + 8 35 + 24 5 + 64 5 + 60 = 40 + 280 + 360 + 320 + 60 = 260. 3
Hece the expected umber of marbles remaiig i bag Y is 260 2 0. The blue marble is of the 5 marbles i bag Y. So the probability that the blue marble remais i bag Y is of the expected umber of marbles 5 remaiig i bag Y : 260 252 63 = = 5 20 20 2 = 63 8 256. Note: The aswer happes to equal ( 0 5 /2 0. That s ot a accidet! See www.aops.com/forum/viewtopic.php?t=499689 for a explaatio. Problem 6 Say that a complex umber z is three-presetable if there is a complex umber w of absolute value 3 such that z = w. Let T be the set w of all three-presetable complex umbers. The set T forms a closed curve i the complex plae. What is the area iside T? Express your aswer i the form a π, where a ad b are positive, relatively-prime itegers. b Aswer: 80 9 π. Solutio: Let w be a complex umber with absolute value 3. Let z be the three-presetable umber w. The z ca be rewritte as w z = w w = w w w 2 = w w 9. Let w = x + yi, where x ad y are real umbers. By the previous equatio, we have z = x + yi x yi 9 = 8 9 x + 0 9 yi. I other words, to go from w to z, stretch the real part by a factor of 8 9 ad stretch the imagiary part by a factor of 0. So T (the set of threepresetable umbers is a ellipse formed by stretchig a circle of radius 3 9 by a factor of 8 i the x-directio ad by a factor of 0 i the y-directio. 9 9 The area iside a circle of radius 3 is 9π. So the area iside T is 8 9 0 80 9π = 9 9 π. 4
Problem 7 How may ordered triples (a, b, c, where a, b, ad c are from the set {, 2, 3,..., 7}, satisfy the equatio Aswer: 408. a 3 + b 3 + c 3 + 2abc = a 2 b + a 2 c + b 2 c + ab 2 + ac 2 + bc 2? Solutio: Let s move all the terms of the equatio to the left side: a 3 + b 3 + c 3 + 2abc a 2 b a 2 c b 2 c ab 2 ac 2 bc 2 = 0. We will try to factor the left side. To do so, let s view the left side as a cubic polyomial i a. We ca group the terms as follows: a 3 (b + ca 2 (b 2 2bc + c 2 a + b 3 + c 3 b 2 c bc 2 = 0. We ca factor the liear coefficiet ad the costat coefficiet: a 3 (b + ca 2 (b c 2 a + (b c 2 (b + c = 0. That helps us partially factor the left side: (a b c [ (b c 2 a 2] = 0. By differece of squares, we ca fully factor: (a b c(b a c(c a b = 0. So either a = b + c or b = a + c or c = a + b. Because the variables are positive, those three cases are mutually exclusive. Hece we ca cout the umber of solutios to a = b + c ad the multiply by 3. Let s cout the solutios to a = b + c, where each of the variables is a iteger from to 7. We have ( 7 2 choices for a ad b. For each such choice, ( we have exactly choice for c. So the umber of solutios to a = b + c is 7 2, or 36. As we said i the previous paragraph, we eed to multiply this cout by 3 to get the fial aswer. Hece the aswer is 3 36, or 408. Problem 8 Sherry starts at the umber. Wheever she s at, she moves oe step up (to 2. Wheever she s at a umber strictly betwee ad 0, she moves oe step up or oe step dow, each with probability. Whe she 2 reaches 0, she stops. What is the expected umber (average umber of steps that Sherry will take? Aswer: 8. 5
Solutio: We ca draw a state diagram to represet Sherry s trip: 2 3 4 5 6 7 8 9 0 Whichever state Sherry is i the diagram, she chooses a out-goig edge at radom as her ext step. The labels represet the cost of makig a step. Sice each cost label is, at each state the expected cost of oe radom step is. Because the problem asks oly about the expected cost of the etire trip, we ca reumber the labels as log as the expected cost of oe radom step remais. Cosider the followig reumberig: 3 5 7 9 3 5 7 2 3 4 5 6 7 8 9 0 3 5 7 9 3 5 At each state, the expected cost of oe radom step is still. For example, at state 4, the out-goig edges are labeled 5 ad 7, for a average of. So we ca aalyze the expected cost of the trip uder the ew umberig. I ay path from to 0 uder the ew umberig, there will be cacellatio. For example, there might be several uses of the 5 ad 5 labels, but there will always be exactly oe more use of the 5 label tha the 5 label. So the total cost will always be + 3 + 5 + 7 + 9 + + 3 + 5 + 7. The sum of these 9 odd umbers is 9 2, or 8. always 8, the expected cost is 8. Because the total cost is Problem 9 Defie L(x = x x2 for every real umber x. If is a positive 2 iteger, defie a by ( ( ( ( 7 a = L L L L, where there are iteratios of L. For example, ( ( ( ( 7 a 4 = L L L L. 4 6
As approaches ifiity, what value does a approach? Express your aswer as a fractio i simplest form. Aswer: 34 9. Solutio: From the defiitio of L, we see that L obeys the followig three properties:. If x 0 ad x 2, the 2. L(x x. 3. If 0 < x < 2, the L(x > 0. L(x x = 2 x. Sice we re iterested i approachig ifiity, we may assume that 9, so 7 is strictly betwee 0 ad 2. If k is a oegative iteger, let L(k be the kth iterate of L. By Property 2, we have L (k ( 7 7 for every k. By Property 3, we have L (k ( 7 > 0 for every k. Combiig these last two iequalities, we have Hece by Property, we have 2 < 2 L (k (7/ 2 7/. 2 < L (k+ (7/ L (k (7/ 2 7/. By summig the previous iequality for all k from 0 to, we have Because a is L ( (7/, we have Dividig by gives 2 < L ( (7/ 7/ 2 7/. 2 < a 7 2 7/. 2 < a 7 2 7/. 7
As approaches ifiity, the left ad right expressios approach. Hece the 2 middle expressio does too. So a approaches + 9, which is. Therefore, 2 7 34 a approaches 34 9. Problem 20 There are 6 distict values of x strictly betwee 0 ad π 2 that satisfy the equatio ta(5x = 5 ta(x. Call these 6 values r, r 2, r 3, r 4, r 5, ad r 6. What is the value of the sum + + + + +? ta 2 r ta 2 r 2 ta 2 r 3 ta 2 r 4 ta 2 r 5 ta 2 r 6 Express your aswer as a fractio i simplest form. Aswer: 78 5. Solutio: We will study the more geeral equatio ta(x = ta(x, where is a fixed iteger greater tha. The multiple-agle formula for taget says that ta x = k=0 ( k( 2k+ (ta x 2k+ k=0 ( k(. 2k (ta x 2k If ta(x = ta(x, the the formula becomes k=0 ta x = ( k( 2k+ (ta x 2k+ k=0 ( k(. 2k (ta x 2k Clearig the fractio, we get ( ( k (ta x 2k+ = 2k k=0 ( ( k (ta x 2k+. 2k + k=0 Brigig all the terms to oe side, we get ( ( [ k 2k k=0 ( ] (ta x 2k+ = 0. 2k + 8
We ca simplify the bracketed expressio usig the biomial idetity ( ( ( + = 2k. 2k 2k + 2k + With this idetity, our mai equatio becomes ( + ( k 2k (ta x 2k+ = 0. 2k + k=0 The k = 0 term is 0 ad so we ca remove it: ( + ( k 2k (ta x 2k+ = 0. 2k + k= Because x is acute, ta x is positive, ad so we ca divide both sides by 2(ta x 3 : ( + ( k k (ta x 2(k = 0. 2k + k= Let t = ta 2 x. The equatio becomes ( + ( k k t k = 0. 2k + k= The costat coefficiet of the left-had side is ( + 3. The liear coefficiet is 2 ( + 5. So by Vieta s formula, the sum of the reciprocals of the roots of t is 2 ( + 5 ( 2( 3 ( + =. 0 Whe = 5, this sum is 3 3 2 0 = 3 6 5 = 78 5. Note: By lettig approach ifiity, we ca show that the squared reciprocals of the positive solutios to ta x = x add up to 0. 9